Question

The sum to $(n+1)$ terms of the series $\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+\cdots$ is

• A. $\frac{1}{n+1}$
• B. $\frac{1}{n+2}$
• C. $\frac{1}{n(n+1)}$
• D. $\frac{1}{(n+1)(n+2)}$

Answer 1

The correct option is D $\frac{1}{(n+1)(n+2)}$

Given:

$\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\cdots +(-1{)}^n\frac{C_n}{n+2}$

Consider the expansion of

$(1-x{)}^n=C_0-C_{1}x+C_2{x}^2-\cdots +(-1{)}^n{C}_n{x}^n$

Multiply both sides with $x$, we get

$x(1-x{)}^n=C_{0}x-C_1{x}^2+C_2{x}^3-\cdots +(-1{)}^n{C}_n{x}^{n+1}$

Integrating both sides w.r.t $x$ from $0$ to $1$,

${\int }_0^{1}x(1-x{)}^{n}dx={\int }_0^1(C_{0}x-C_1{x}^2+C_2{x}^3-\cdots +(-1{)}^n{C}_n{x}^{n+1})dx$

$\Rightarrow {[\frac{(1-x{)}^{n+2}}{n+2}-\frac{(1-x{)}^{n+1}}{n+1}]}_0^1={[\frac{C_0{x}^2}{2}-\frac{C_1{x}^3}{3}+\frac{C_2{x}^4}{4}-\cdots +(-1{)}^n\frac{C_n{x}^{n+2}}{n+2}]}_0^1$

$\Rightarrow \frac{1}{(n+1)(n+2)}=$$\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\cdots +(-1{)}^n\frac{C_n}{n+2}$

Hence, option D.